1. Aug 26, 2011

### pjpell

This is probably really simple for many here, but I have not done physics since high school. I need help settling a huricane-irene induced bet...

You stand in place a throw a golf ball straight up in the air such that the total hang time (if there was no wind) was exactly 5 seconds. Now let's say there is constant wind in one direction at 67 miles/ hour. How far away would the ball land?

2. Aug 26, 2011

### Hootenanny

Staff Emeritus

Welcome to Physics Forums.

Well, assuming that the wind is blows horizontally, you threw the ball to the same height and the ground was level, the ball's flight time wouldn't change. 67 mph is about 30 m/s. Therefore, the ball would land $30\times 5 = 150$ meters away.

3. Aug 26, 2011

### potatoecannon

150m is a bit of a simplification; this assumes the ball is instantly accelerated to 30m/s horizontally when it leaves your hand.

The actual distance would be slightly less, but for the purposes of a bet it's close enough.

4. Aug 26, 2011

### jambaugh

Problem is that the 30m/s calculation assumes the ball immediately moves horizontally with the wind. The actual horizontal flight will be less than 150meters since for most of the ball's flight it will be moving horizontally slower than the wind.

To solve the problem exactly you need to model the drag on the ball due to wind. This is not a simple task but you could use a simplified Euler drag model, the ball's velocity relative to the wind will decay exponentially. In that case you still must determine empirically the drag coefficient.

Assume that the ball's velocity relative to the wind is q times the windspeed (q<1) after 5 seconds or relative to the ground p times the windspeed (p=1-q).
You have a velocity relative to the wind of $\Delta v(t) = v_w q^{t/5}$
The velocity relative to the ground is then:
$v(t) = v_w(1-q^{t/5})$

Integrating with x(0)=0 yields: $x(t) = v_w(t + \frac{5}{\ln(q)}[1- q^{t/5}])$

At t=5 you get $x(5) = v_w(5 + \frac{5(1-q)}{ln(q)})$.
For windspeed 30m/s that's:
$x(5) = 150(1 + \frac{(1-q)}{\ln(q)})= 150(1+\frac{p}{\ln(1-p)})$

Some typical values in meters:
p-----x(5)
0.999999999999999-----145.6571557
0.999-----128.3069906
0.9-----91.37024494
0.8-----75.43980785
0.7-----62.78872777
0.6-----51.77789989
0.5-----41.79787193
0.4-----32.54308866
0.3-----23.83470366
0.2-----15.55739647
0.1-----7.631676285
0.0001-----0.007500125

5. Aug 26, 2011

### Hootenanny

Staff Emeritus

As has been point out, I should have added the caveat that it was only a rough back of a napkin calculation. I thought a rough answer would be sufficient for the purposes of a bet.

6. Aug 26, 2011

### Neandethal00

150m is correct because I think OP was asking for an answer in an ideal case. There is no horizontal acceleration or deceleration. An ideal constant 30m/sec speed of the ball the moment it left the hand.

If you want to be perfect and take air drag into account, you have to understand that a constant 30m/sec wind velocity throughout its flight is virtually impossible. Then a gradient of wind velocity v(h), h is height, will also enter into the formula.

7. Aug 26, 2011

### cjl

I would disagree that 150m is correct for an ideal case, because I wouldn't consider an infinite drag coefficient in the horizontal direction, but zero in the vertical direction to be ideal (or accurate). I used my matlab ballistic trajectory calculator (I wrote it a while ago for projectiles with drag), and I used an initial vertical velocity to provide a 5 second hangtime with no aerodynamic drag. I then added in aerodynamic drag using a drag coefficient of 0.3 (which is about right for a sphere in turbulent flow), and I get the following:

Hangtime: around 4.5 seconds
Downrange distance: around 32 meters

8. Aug 26, 2011

### MrREC

Sorry for butting in but......... How high would you have to throw the ball to get a 5 second "hang time"? 75ft or more?

9. Aug 26, 2011

### cjl

My simulation for the 4.5 second hang time (with drag) shows around 25 meters.

10. Aug 26, 2011

### MrREC

There seems to be some differences in the answers to the question.

That being said ..... The best test would be to accelerate the ball to 67 mph, and then have it free-fall drop for 5 seconds in a vacuum. Since there is no way to do that test, that I am aware of, then the answer would have to vary from the shortest to the longest (32m to 150m)?

Plus if you discount point of impact, and consider total distance traveled after impact, then the distances will vary even more. (impact on sand versus a frozen lake could vary quite a bit in the total distance traveled).

( .....just my two cents)

11. Aug 26, 2011

### cjl

I don't see how that's a good test - that will just give the 150m result (a fact which is trivially easy to show without an experiment). However, the question was about an initially stationary ball thrown up in a windy condition, which will give a substantially different result.

12. Aug 26, 2011